Expansion irreversibly and isothermally of an ideal has from a volume of 2dm3 to 10 dm3 against a const pressure of 1 atm at 27 C .Find entropy change of the system and also surroundings due to the process.
Heres my attempt:.....
delta S=energy absorbed/temp........
for the system,......
= energy absorbed/temp=RTln(V2/V1)/T(as it is an isothermal exp)........
putting in values I got,13.4J/K/mol......
since energy absorbed by surroundings= -(energy absorbed by system).....
delta S of surroundings= -13.4J/K/mol......
My book gives different answer fro 2nd part....
Why am I wrong?...
I just don't understand when we have to consider work and when not?(like for the system or the surroundings?)....
Entropy change = Qrev / T formula is applicable when the process is reversible. A common way to find entropy change for irreversible process is to define new reversible process with the same end points and find entropy change for that.
I have not seen in any book that it is applicable only for reversible process.It has been given that when He and kr are in two different chmbers connected by a tube fitted with piston,then entropy is less.And when the piston is removed the gases are mixed with each other.Is it not an irreversible process?This example has been given in book about entropy.